bambi462
bambi462
04-09-2014
Mathematics
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(3x-4)∧2=8 how do i solve this
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Аноним
Аноним
04-09-2014
[tex] (3x-4)^2=8\iff3x-4=\pm\sqrt8\\\\3x-4=-\sqrt{4\times2}\ or\ 3x-4=\sqrt{4\times2}\\\\3x-4=-2\sqrt2\ or\ 3x-4=2\sqrt2\ \ \ \ \ \ \ |add\ 4\ to\ both\ sides\\\\3x=4-2\sqrt2\ or\ 3x=4+2\sqrt2\ \ \ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{x=\frac{4-2\sqrt2}{3}\ or\ x=\frac{4+2\sqrt2}{3}}[/tex]
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